Lecture Notes in Differential Equations - Bruce E. Shapiro

370 LESSON 33. NUMERICAL METHODS
Heun’s Method is
y n = y n−1 + h n
4
[
f(t n−1 , y n−1 ) + 3f
(
t n−1 + 2 3 h, y n−1 + 2 )]
3 hf(t n−1, y n−1 )
(33.67)
Both Heun’s method and the modified Euler method are second order and
are examples of two-step Runge-Kutta methods. It is clearer to implement
these in two “stages,” eg., for the modified Euler method,
while for Heun’s method,
ỹ n = y n−1 + hf(t n−1 , y n−1 ) (33.68)
y n = y n−1 + h n
2 [f(t n−1, y n−1 ) + f(t n , ỹ n )] (33.69)
ỹ n = y n−1 + 2 3 hf(t n−1, y n−1 ) (33.70)
y n = y n−1 + h [
(
n
f(t n−1 , y n−1 ) + 3f t n−1 + 2 )]
4
3 h, ỹ n (33.71)
Runge-Kutta Fourth Order Method. This is the “gold standard” of
numerical methods - its a lot higher order than Euler but still really easy
to implement. Other higher order methods tend to be very tedious – even
to code – although once coded they can be very useful. Four intermediate
calculations are performed at each step:
Then the subsequent iteration is given by
k 1 = hf(t n , y n ) (33.72)
k 2 = hf(t n + .5h, y n + .5k 1 ) (33.73)
k 3 = hf(t n + .5h, y n + .5k 2 ) (33.74)
k 4 = hf(t n + h, y n + k 3 ) (33.75)
y n+1 = y n + 1 6 (k 1 + 2k 2 + 2k 3 + k 4 ) (33.76)
Example 33.2. Compute the solution to the test equation y ′ = y, y(0) = 1
on [0, 1] using the 4-stage Runge Kutta method with h = 1/2.
Since we start at t = 0 and need to compute through t = 1 we have to
compute two iterations of RK. For the first iteration,
k 1 = y 0 = 1 (33.77)